Tread Mills

General
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Treadmills

I hope no one thinks I’m being argumentative here, 'cause that’s not my intent. This is just an interesting problem, and I’m enjoying talking about it.

I’m not convinced that this analogy fits when it comes to ascending on a unicycle, or b***, for that matter, because the circle that your knees are making as you pedal doesn’t change in size whether you’re riding on the flat or uphill.

The pull of gravity doesn’t change while you are in contact with the surface. As long as you aren’t hopping up the hill, the potential energy is constant. Given an ideal, even grade, you won’t have to work harder to keep going up at a higher altitude than at a lower one. The amount of energy required to go up will remain constant. If you started at the bottom, and are halfway up the incline, you’re going to be tired, and so it may feel harder, but that’s not from potential energy, it’s from lactic acid.

I don’t think that potential energy completely explains why it’s more difficult to go uphill than to ride on a level surface, because as long as you’re in contact with the surface, there’s nothing that requires you to expend more energy relative to altitude on an unchanging grade.

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Re: Treadmills

Are you sure you’re not confusing gravity (force) and energy? While gravity will stay the same whether you’re at the bottom of the hill, the potential energy increases as you go upwards.

Going back to U=mgh… this is the energy of a falling body; the potential energy is where that energy comes from when it starts to fall.

Phil, just me

43

Treadmills

Yes, but upwards relative to what?

I always thought that meant altitude above a surface, as in “in mid-air”. You gain potential energy when you leave the surface.
If you’re on the surface, how can the potential energy increase? Please don’t tell me anymore disappearing mountain stories. If you’re standing on the summit of Mt. Everest, and you drop your oxygen canister onto your foot, it’s not going to hurt any more or less than if you dropped the same canister onto your foot down on the beach in Phuket (I don’t know any sea-level location names in Tibet).

Well, OK, you’ll have to be wearing the same triple-insulated plastic mountaineering boots in both places as a control-factor.

I thought you said your brother was right?

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Re: Treadmills

Upwards relative to an arbitrary point you choose. Technically you’ve only got no potential energy if you are completely weightless, for example in the middle of the earth*.

How much potential energy you’ve got in total isn’t the point, it’s how much it changes between one altitude and another.

For Harper’s equation U=mgh U is the change in Ep, not the absolute value of Ep (unless you’re measuring h from the centre of gravity of the earth).

Going further upwards means you’re going away from the centre of gravity of the earth, and so gaining potential energy.

For your dropping of an oxygen canister example, it’d hit your foot at the same speed because the change in potential energy is the same in both cases.

(My brother was right about having to do work on an inclined treadmill; I initially argued that as you’re not actually going upwards it wouldn’t be any different than on the flat… but as the thing would otherwise pull you downwards you’ve got to fight against that)

Phil, just me

  • PEDANTS: Yes, I know, the centre of the universe of wherever. Stop being picky…
45

Treadmills

My point is that altitude is irrelevant as long as you’re touching the ground. Just because you’re 100 feet up along an incline doesn’t mean you’re going to fall all the way to the bottom. The potential energy change is illusory (as pertaining to your potential to become a falling body and accelerate downwards) unless you’re ascending a very steep grade. That’s what the oxygen cannister thing was meant to illustrate.

For the sort of grade on which it is reasonably possible to maintain upward motion (because you have enough friction to avoid sliding backwards) on a unicycle, if you travel upwards from 5 feet to 25 feet above sea level, or from 500 feet to 525 feet above sea level, given the same grade, you’re going to use the same energy.

Sal said it felt like he was going uphill.

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Re: Treadmills

(I’m assuming you meant 25 feet in both cases; 25-5 isn’t 525-500…)

In both cases you’re going up 25 feet, so yes it uses the same amount of energy to move in the vertical plane; 25 x gravity x your mass.

If you were just going along the flat, you’d not be moving in the vertical plane, so you’d not use any energy, and so not gain any potential energy.

Phil, just me

47

Treadmills

Yes, that was sloppy of me, wasn’t it? Thanks for kindly giving me the benefit of the doubt! :slight_smile:

But whoa! No energy to move along a flat surface? That’s good news!

Or bad news for all those stationary joggers who are trying to slim down. :wink:

Momentum.

That’s what this thread’s got.

Been fun, but gotta go…

Glutes

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Re: Treadmills

Yes, this thread is getting rather long… :slight_smile:

No energy required to move in a vertical direction when going along a level surface. I didn’t say anything about going horizontally…

Phil, just me

49

You’ve probably finished this debate, but that’s a lot of posts that I’ve read in one go that I’ve wanted to talk about, so I’m going to have to catch up a bit.

Firstly, potential energy can be understood easier if you consider a ball (for example) rolling down a hill. At the top of the hill it has high potential energy, and no kinetic energy. (I’m using sea level as a reference here. It makes sense, 'cause it can be sued for pretty much every situation. Unless anyone unicycles underwater. But in that case you have buoyancy as well, which confuses everything. And I’m talking about kinetic energy as well, because it helps. Honestly! Anyway, I had a point once, what was it?) Right. If you let the ball go from the top of the hill, it rolls down the hill. It clearly gains kinetic energy, because it speeds up. Energy cannot be created or destroyed, so the kinetic energy has to have come from somewhere. In this case it has come from the potential energy. The increase in kinetic energy is equal to the loss of potential energy.

The main point in my original post was that on a treadmill you don’t gain potential energy or kinetic energy, but you have to do work against the motor, which is moving the belt in a way that would pull you downwards, thereby reducing your potential energy, if you stopped pedalling. On a treadmill, you aren’t doing work to increase your potential energy, you’re doing the same amount of work to stop it being decreased.

The other bit was about travelling in a horizontal direction. Here, finally, we can forget potential energy. Here, newton’s law of force = mass * acceleration is what you want. That means a force is only required to change the speed of something. So in horizontal travel, you exert a force when you start to increase your speed, but then stop exerting the force when you reach a constant speed. “But it bloomin’ well feels like I’m exerting a force”, I hear you shout. This is true, because you have a constant frictional force opposing you. The force in the newton’s thing is actually resultant, or overall, force. The force you exert at a constant speed is equal to friction, so that the total force on you / your unicycle is 0, and there is no acceleration or decceleration. Another way of looking at it is that if there was no friction, you wouldn’t have to exert a force to travel at constant speed. For example, if you unicycled on ice, with slick tyres (which would be pretty impressive) the friction force would be tiny, and you would hardly have to do any work to keep going. However you wouldn’t be able to go anywhere, because apart from the balance thing friction is the only way you can exert a force on the surface your riding on to speed up ort slow down.

Right, I’m stopping now. That’s a bloomin’ long post. I really need to get out more.

P.S. Just to confuse things (even more), the energy cannot be created or destroyed thing falls apart with quantum physics, where you have a scary complicated mass / energy duality thing, but that’s far too complicated, and isn’t particularly relevant. Unless anyone’s considering a fusion powered unicycle. I didn’t really need to say this, but I don’t want anyone else to beat me to it. Not that anyone would do that. Particularly anyone doing computing at york uni. Never.

50

Umm…what he said. I agree johnhimsworth. Are you still saying that they act the same? The tread mill and the hill?

51

Yes, I’m still saying you have to put in just as much effort. A treadmill might be a bit easier because of having a smooth, grippy surface (look! Grippy! That might not even be a word, let a alone a scientific word. For those upset by that, substitute something about a high coefficient of frcition) and because you aren’t moving there won’t be any air resistance, although I don’t think air resistance going uohill on a uni is that strong.

That was a bit of a heavy post, wasn’t it. Maybe I should write a physics textbook someday. Oh, wait, I just have done. Glad you agree though.

52

There is an increase in potential energy when running on an inclined treadmill; as you put you front foot forward onto the higher surface, you are gaining potential energy; the same amount of potential energy that you would if you were placing it to a higher point on an actual hill. The difference is that on a treadmill it’s immediately converted into kinetic energy as it pushes the treadmill backwards.

I haven’t done a vector diagram, but I expect that the amount of energy required to stay in one place on an inclined treadmill is equivalent to the amount of energy required to walk up a smooth, inclined surface.

Actually, I would say it takes more energy on the treadmill, because it’s so friggin’ boring.

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But at the same time your other leg is going backwards, going further down the hill, losing potential energy. Think centre of mass. It doesn’t move.

Why the dragging from the archive anyway?

John

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This thread was referenced in another post.

Obviously, when you’re walking/riding on a treadmill your overall potential energy doesn’t change, but I would suppose that is only because it is being immediately turned into kinetic energy. The potential energy of your forward foot rises as you move it up the slope, and then dissipates as it pushes the treadmill fabric backwards.

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This nerdy thread is largely ignoring the dynamics of the human body. The formulae and arguments I’m reading are mostly related to point masses / rigid bodies, and the human body shouldn’t be reduced to a point mass in this case. Ferchrissakes, model the problem correctly!

Granted, kinetic and potential energies are involved, but to a human being the quality that is most directly observed is “perceived work” (i.e. energy expended) when engaged in an activity such as running / riding, on or off a treadmill. Thinking about the center of mass is only useful in determining quantities such as ground speed (which is zero for one on a treadmill, quite boring!), net acceleration and drag forces (again, zero).

If one is to attempt to calculate the actual quantity of work done, than each segment of the human body (i.e. torso, thigh, calf, foot, etc.) needs to be considered separately, as well, coupling equations (involving driving and frictional forces) must be considered for the intersegment connections (a.k.a joints & ligaments). This can get very complicated, very quickly: the hip lifts the leg, the thigh extends the calf and foot. Work must be done to overcome the inertia of the thigh, calf and foot, the friction in the joints, as well as gravity / body weight. Etc, etc, etc.

Some of the earliest posts in this thread had made similar points in this regard.

Sure, more work will need to be done as the grade is increased or decreased from the horizonal, but this isn’t a surprise. No complicated mathematics are required! If one thinks about this problem in the frame of reference of the treadmill’s belt, then the problem becomes much easier An inclined treadmill = a hill, while a level (horizontal) treadmill = flat ground. This is your first-order approximation. 'Nuff said.

This talk of motors and friction doesn’t relate to the rider on the treadmill! The rider is merely overcoming gravitational forces (again, think frame of reference), while the motor is doing the work to overcome the load of the rider: to the motor, the load is created in the friction between the belt and the belt support, which is increased by the weight of the rider. You can put all the friction you want (or don’t want) between the belt and its support; the rider won’t know the difference!

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No one in their right mind would make this statement because it is absolutely incorrect (and irrelevant to boot, who’d want to “beat you” to making such a blunder?) There’s nothing complicated about the phenomenon you’re trying to obfuscate: Einstein’s famous formula E=mc^2 says it all (when generalized, see below). Energy is never created or destroyed! Furthermore, there is no duality: one can observe (or deduce) the quantity of mass a body contains, how could it be mistaken as energy?

What you allude to is the fact that energy can change forms, and this isn’t complicated either (as you illustrated earlier in your same post). If potential energy can be simply converted into kinetic energy (by rolling a ball down a hill), why is it so hard to imagine that kinetic energy can be converted into mass?

Einstein’s “famous formula” is generalizable into an energy-momentum four-vector as E^2 = (mc^2)^2 + (pc)^2 (where p is a momentum vector; other generalizations exist, but this best illustrates my point). From this one can “do the math” and verify any mass - energy conversion! This is simple algebra, folks, multiplication and addition, we can all do this!

John, I hereby revoke your license to use the phrase “quantum physics” until you’ve refreshed your knowledge of the subject. Until such refreshment, all future use will be considered illegal and subject to further flaming. I recommend the text “Modern Physics” by Paul Tipler and Ralph Lewellyn.

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Please send pictures! The one I have is from 1981 and is on an un-powered treadmill. I want the real thing!

Someday that page will be brought into the 21st century!
http://www.unicycling.com/things/

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More followup:

Looks like we have no firsthand examples for people to report since 2002. Too bad. I missed my chance to play on a motorized treadmill a couple of times since then. Usually there’s not a unicycle around when you have a powered treadmill handy, and nobody watching.

Hypothesis:

  • Since it greatly complicates the mechanism, adds parts and greatly increases the cost of the device, I submit that tilting a treadmill works. Somebody prove me wrong; good luck.

  • I will also guess, with less certainty, that riding a unicycle on a tilted treadmill will feel like going uphill, at least a little bit. The difference here will be that there is no lifting of the feet to a higher spot. However I believe the rider’s balance envelope will be shrunken just as it is on a real hill, making it a little harder to ride the angle. Also you will have to put some degree of additional energy into the wheel to resist its tendency to roll backwards.

But I don’t think it will feel like as much additional work as it would for someone walking/running.

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I disagree; I think riding a unicycle on an inclined treadmill requires pretty much the same amount of effort as riding up a smooth incline, except for overcoming wind resistance. Riding uphill takes more work because you are being accelerated down the hill by gravity; that will happen in the same degree on an inclined treadmill.

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Re: Tread Mills

“maestro8” <maestro8@NoEmail.Message.Poster.at.Unicyclist.com> writes:

> johnhimsworth wrote:
> > *P.S. … the energy cannot be created or destroyed thing falls
> > apart with quantum physics
>
> No one in their right mind would make this statement because it is
> absolutely incorrect (and irrelevant to boot, who’d want to “beat you”
> to making such a blunder?) There’s nothing complicated about the
> phenomenon you’re trying to obfuscate: Einstein’s famous formula E=mc^2
> says it all (when generalized, see below). Energy is never created or
> destroyed!

E=mc^2 is a classical result which may break from a quantum mechanical
perspective. How do you explain virtual particles that, according to
quantum theory, act as intermediates in interactions at the quantum
level? What about the virtual particles which form just outside of
the event horizon of black holes, become real, and escape as Hawking
radiation?

I’m no expert on quantum physics, but it seems to me that many
reasonable people might describe quantum violations of the first law
of thermodynamics.

Looking forward to your response,

Ken